Question: $g(x)=\begin{cases} \dfrac x3-2&\text{for }0<x<6 \\\\ \text{cos}(x\cdot\pi)&\text{for }6\leq x\leq10 \end{cases}$ Find $\lim_{x\to 6}g(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $0$ (Choice C) C $1$ (Choice D) D The limit doesn't exist.
Solution: $x=6$ is on the boundary between the pieces of our piecewise function. In order to find $\lim_{x\to 6}g(x)$, we need to find the one-sided limits. Let's find the limit as $x$ approaches $6$ from the left. We will use the fact that $g(x)=\dfrac x3-2$ for $x$ -values smaller than $6$. $\begin{aligned} &\phantom{=}\lim_{x\to 6^-}g(x) \\\\ &=\lim_{x\to 6^-}\dfrac x3-2 \\\\ &=\dfrac63-2&\gray{\text{Direct substitution}} \\\\ &=0 \end{aligned}$ Let's find the limit as $x$ approaches $6$ from the right. We will use the fact that $g(x)=\text{cos}(x\cdot\pi)$ for $x$ -values greater than $6$. $\begin{aligned} &\phantom{=}\lim_{x\to 6^+}g(x) \\\\ &=\lim_{x\to 6^+}\text{cos}(x\cdot\pi) \\\\ &=\text{cos}(6\pi)&\gray{\text{Direct substitution}} \\\\ &=1 \end{aligned}$ $0\neq 1$ so the one-sided limits aren't equal. This means that the two-sided limit $\lim_{x\to 6}g(x)$ doesn't exist.